One Faraday of electricity is pa ssed through molten `Al_(2)O_(3)`, aqeusous solution of `CuSO_(4)` and molten NaCl taken in three different electrolytic cells connected in seris. The mole ratio of Al, Cu,Na deposted at the respective cathode is

Options

A. `2 : 3 : 6`
B. `6 : 2 : 3`
C. `6 : 3 : 2`
D. `1 : 2 : 3`

Answer

A

Explanation

`Al^(3+) + 3e^- to Al`
`Cu^(2+) + 2e^- to Cu`
`Na^+ + e^- to Na`
Thus , 1 F of electricity will deposit `1/3` mol of Al `1/2` mol of Cu and 1 mol of Na . Hence , mole ratio of Al , Cu , Na deposited at respected cathodes is :
`1/3 : 1/2 : 1 implies 2 : 3 : 6`